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A Family of Tetravalent Half-Transitive Graphs

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A Family of Tetravalent Half-Transitive Graphs A Family of Tetravalent Half-transitive Graphs Sucharita Biswas1, Angsuman Das1,∗ Department of Mathematics, Presidency University, Kolkata, India Abstract In this paper, we introduce a new family of graphs, Γ(n, a). We show that it is an infinite family of tetravalent half-transitive Cayley graphs. Apart from that, we determine some structural properties of Γ(n, a). Keywords: half-transitive graph, graph automorphism, cycles 2008 MSC: 05C25, 20B25, 05E18 1. Introduction A graph G =(V, E) is said to be vertex-transitive, edge-transitive and arc-transitive if the automorphism group of G, Aut(G), acts transitively on the vertices, on the edges and on the arcs of G respectively. It is known that an arc-transitive graph is both vertex-transitive and edge-transitive. However, a graph which is both vertex-transitive and edge-transitive may not be arc-transitive, the smallest example being the Holt graph [9] on 27 vertices. Such graphs are called half-transitive graphs. For other definitions related to algebraic graph theory, one is referred to [8]. The study of half-transitive graphs was initiated by Tutte [12], who proved that any half-transitive graph is of even degree. Since any connected 2-regular is a cycle and a cycle is arc-transitive, the first possibility of finding a half-transitive graph is a 4-regular or tetravalent graph. The first example of tetravalent half-transitive graphs were given by Bouwer [2] and the smallest example was given in Holt [9]. Though numerous papers have arXiv:2008.07525v1 [math.CO] 16 Aug 2020 been published in the last 50 years, the classification of tetravalent half-transitive graphs is not yet complete. In absence of complete classification, two major approaches have been fruitful so far: the first is to characterize half-transitive graphs of some particular orders like p3,p4,p5, pq, 2pq etc [4],[5],[6],[7], and the second is to come up with infinite families of half-transitive graphs [3],[13]. In this paper, we introduce a new family of tetravalent half-transitive graphs, Γ(n, a). It turns out that it is family of Cayley graphs of order 3n and for (n, a) = (9, 4), we get ∗Corresponding author Email addresses: [email protected] (Sucharita Biswas), [email protected] (Angsuman Das) Preprint submitted to August 19, 2020 the Holt graph. In fact, our construction is a generalization of an alternative construction of Holt graph (See the last paragraph of [9]). Definition 1.1. Let n be a positive integer such that 3 | ϕ(n), where ϕ denote the Euler Z∗ Z Totient function. Then n, the group of units of n, is a group of order a multiple of 3. Z∗ 2 Let a be a element of order 3 in n and b = a (mod n). Define Γ(n, a) to be the graph with vertex-set Zn × Z3 and the edge-set comprises of edges of the form (i, j) ∼ (ai ± 1, j − 1) and (i, j) ∼ (bi±b, j +1), where the operations in the first and second coordinates are done in modulo n and modulo 3 respectively. It is obvious that Γ(n, a) is tetravalent. One can check that Γ(9, 4) is the Holt graph. It is also to be noted that for a particular n, we can have two graphs, Γ(n, a) and Γ(n, b). However, these two graphs are isomorphic via the automorphism τ : Γ(n, a) → Γ(n, a2) defined by τ(i, j)=(ai, −j). So, without loss of generality, we assume that a < b, where a, b ∈{2,...,n − 2}. On the other hand, let n be a positive integer such that a1, b1, a2, b2 be four elements of Z∗ order 3 in n with a1b1 ≡ 1(mod n) and a2b2 ≡ 1(mod n). Then, by the above argument, Γ(n, a1) ∼= Γ(n, b1) and Γ(n, a2) ∼= Γ(n, b2). However, Γ(n, a1) may not be isomorphic to Γ(n, a2), e.g., for n = 63, we have 4·16 ≡ 1(mod 63) and 22·43 ≡ 1(mod 63), but Γ(63, 4) is not isomorphic to Γ(63, 22), as odd girth of Γ(63, 4) is 9, whereas that of Γ(63, 22) is 21. The definition of Γ(n, a) requires that 3|ϕ(n). We discuss the form of n, for which this α1 α2 αk α1−1 α2−1 αk−1 holds. Let n = p1 p2 ··· pk , wher pi are primes. Then ϕ(n)= p1 p2 ··· pk (p1− αi−1 1)(p2 − 1) ··· (pk − 1). As 3|ϕ(n), either 3|pi or 3|pi − 1 for some i, i.e., 9|n or pi ≡ 1 (mod 3) for some i. Thus n is either of the form 9t or pt where p is a prime of the form 1 (mod 3) and t is a positive integer. At this junction, it is important to mention that authors in [1] constructed a family of graphs M(a; m, n) and proved the following theorem. Theorem 1.1 ([1], Theorem 3.3). Let n ≥ 9 be odd and a3 ≡ 1 (mod n). Then M(a;3, n) is half-transitive. While the current work was in progress, it caught our attention that our construction Γ(n, a), coincidentally, is same as M(a;3, n) (Although the definitions are given in com- pletely different way). However for two reasons we decided to carry on with the project: 1. Firstly, the proof techniques are entirely different: While their proof is built on semiregular automorphisms and blocks, ours is based on 6-cycles present in the graph. 2. Secondly and most importantly, we prove that Γ(n, a) is half-transitive for all n except 7 and 14, i.e., n is not necessarily odd (See Theorem 1.1). In other words, comparing with Theorem 1.1 we prove the half-transitivity of a larger family of graphs. In Section 2, we discuss some structural properties of Γ(n, a) and in Section 3, we discuss the parameters related to the automorphism group of Γ(n, a). In Section Appendix, we provide the SageMath [11] code for computing the automorphism group of Γ(n, a). 2 2. Structural Properties of Γ(n,a) Theorem 2.1. Γ(n, a) is bipartite if and only if n is even. Proof: If n is even, then consider the sets X = {(i, j): i is even } and Y = {(i, j): Z∗ i is odd}. Note that as n is even and a, b ∈ n, therefore a, b are odd. Thus, if i is even, then ai ± 1 and bi ± b are odd, and if i is odd, then ai ± 1 and bi ± b are even. Thus, no two vertices in X (or Y ) are adjacent and hence Γ(n, a) is bipartite. On the other hand, if n is odd, as Γ(n, a) is Hamiltonian (See Corollary 3.1), it has an odd cycle of length 3n. Hence Γ(n, a) is not bipartite. Theorem 2.2. Then chromatic number χ of Γ(n, a) is given by 2, if n is even χ = 3, if n is odd Proof: The proof is obvious, if n is even. If n is odd, consider the sets A0 = {(i, 0) : i ∈ Zn}, A1 = {(i, 1) : i ∈ Zn} and A2 = {(i, 2) : i ∈ Zn}. The theorem is true as A0, A1 and A2 are independent sets in Γ(n, a) and Γ(n, a) is not bipartite. Lemma 2.1. If n = p, where p ≡ 1 (mod 3) is a prime, then girth of Γ(n, a) is 3. Proof: As a3 ≡ 1(mod n) and a 6≡ 1(mod n), we have n|a3 − 1=(a − 1)(a2 + a + 1). As n is prime, we have a2 + a +1 ≡ 0(mod n). The lemma follows from the observation that (0, 0) ∼ (1, 2) ∼ (a +1, 1) ∼ (a2 + a +1, 0) = (0, 0) is a 3-cycle. Lemma 2.2. Γ(n, a) always have a cycle of length 6 and hence girth of Γ(n, a) is less than or equal to 6. Proof: Consider the cycle (0, 0) ∼ (1, 2) ∼ (a − 1, 1) ∼ (a2 − a +1, 0) ∼ (−a2 + a, 2) ∼ (b, 1) ∼ (0, 0). Clearly it is a 6-cycle, provided the vertices are distinct. If two vertices are not distinct, then we have either a − 1 ≡ b or −a2 + a ≡ 1 or a2 − a +1 ≡ 0, i.e., in all cases, we should have a2 − a +1 ≡ 0 (mod n), i.e., a3 +1 ≡ 0 (mod n). This implies 2 ≡ 0 (mod n), a contradiction. Lemma 2.3. Γ(n, a) does not have any 4-cycle. Proof: Let, if possible, Γ(n, a) has a 4-cycle. As Γ(n, a) is edge-transitive (by Theorem 3.3), without loss of generality, we can assume that it has (0, 0) and (1, 2) as two of its adjacent vertices. Now, the other three neighbours of (0, 0) are X = {(b, 1), (−b, 1), (−1, 2)} and that of (1, 2) are Y = {(2b, 0), (a+1, 1), (a−1, 1)}. To reach a contradiction, it suffices to show that X and Y does not have any edge between them. We start with (2b, 0) ∈ Y . If (2b, 0) ∼ (−1, 2), then by the adjacency criterion, we have 2ab ± 1 ≡ −1 (mod n), i.e., −1 ≡ 3, 1 (mod n), i.e., n divides 2 or 4, a contradiction. If (2b, 0) ∼ (b, 1), then by the adjacency criterion, we have ab ± 1 ≡ 2b (mod n), i.e., 2b ≡ 0, 2(mod n).
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